which are the solutions of $x^{2}=19x + 1$?\n$\frac{19-sqrt{19}}{2},\frac{19+sqrt{19}}{2}$\n$\frac{19-sqrt{36…

which are the solutions of $x^{2}=19x + 1$?\n$\frac{19-sqrt{19}}{2},\frac{19+sqrt{19}}{2}$\n$\frac{19-sqrt{365}}{2},\frac{19+sqrt{365}}{2}$\n$\frac{-19-sqrt{19}}{2},\frac{-19+sqrt{19}}{2}$\n$\frac{-19-sqrt{365}}{2},\frac{-19+sqrt{365}}{2}$

which are the solutions of $x^{2}=19x + 1$?\n$\frac{19-sqrt{19}}{2},\frac{19+sqrt{19}}{2}$\n$\frac{19-sqrt{365}}{2},\frac{19+sqrt{365}}{2}$\n$\frac{-19-sqrt{19}}{2},\frac{-19+sqrt{19}}{2}$\n$\frac{-19-sqrt{365}}{2},\frac{-19+sqrt{365}}{2}$

Answer

Explanation:

Step1: Rewrite the equation in standard form

$x^{2}-19x - 1=0$

Step2: Identify coefficients for quadratic formula

For $ax^{2}+bx + c = 0$, here $a = 1$, $b=-19$, $c=-1$

Step3: Apply the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$x=\frac{-(-19)\pm\sqrt{(-19)^{2}-4\times1\times(-1)}}{2\times1}=\frac{19\pm\sqrt{361 + 4}}{2}=\frac{19\pm\sqrt{365}}{2}$

Answer:

$\frac{19 - \sqrt{365}}{2},\frac{19+\sqrt{365}}{2}$